3.157   ODE No. 157

$$\boxed{{\displaystyle (x^2-1)\frac{\mathrm{d}}{\mathrm{d}x}y\left(x\right)+a({\left(y\left(x\right)\right)}^2-2xy\left(x\right)+1)=0}}$$

1. Problem in Latex
2. Mathematica input
3. Maple input

Mathematica: cpu = 0.086011 (sec), leaf count = 158 $$\left\{\{y(x)\to \frac{(x^2-1)(c_1(ax{(x^2-1)}^{\frac{a}{2}-1}{P}_{a-1}(x)+{(x^2-1)}^{\frac{a}{2}-1}(a{P}_a(x)-ax{P}_{a-1}(x)))+ax{(x^2-1)}^{\frac{a}{2}-1}{Q}_{a-1}(x)+{(x^2-1)}^{\frac{a}{2}-1}(a{Q}_a(x)-ax{Q}_{a-1}(x)))}{a(c_1{(x^2-1)}^{a/2}{P}_{a-1}(x)+{(x^2-1)}^{a/2}{Q}_{a-1}(x))}\}\right\}$$

Maple: cpu = 0.188 (sec), leaf count = 231 $$\{y\left(x\right)=\frac{1}{(4+4x)a}(8((a-1/2)x-a/2+1/2)(1+x)\mathit{\_}\mathit{C}\mathit{1}\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(0,-2a+1,0,0,a^2-a+1/2,2{(1+x)}^{-1})-a{(-\frac{x}{2}-\frac{1}{2})}^{-2a+1}(1+x)\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(0,2a-1,0,0,a^2-a+\frac{1}{2},2{(1+x)}^{-1})-8(x-1)(\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}\mathit{P}\mathit{r}\mathit{i}\mathit{m}\mathit{e}(0,-2a+1,0,0,a^2-a+1/2,2{(1+x)}^{-1})\mathit{\_}\mathit{C}\mathit{1}-1/4{(-x/2-1/2)}^{-2a+1}\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}\mathit{P}\mathit{r}\mathit{i}\mathit{m}\mathit{e}(0,2a-1,0,0,a^2-a+1/2,2{(1+x)}^{-1}))){(\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(0,-2a+1,0,0,a^2-a+\frac{1}{2},2{(1+x)}^{-1})\mathit{\_}\mathit{C}\mathit{1}-\frac{1}{4}{(-\frac{x}{2}-\frac{1}{2})}^{-2a+1}\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(0,2a-1,0,0,a^2-a+\frac{1}{2},2{(1+x)}^{-1}))}^{-1}\}$$